a.
To draw:
The map that shows relative positions of markers and origins of transfer in HfrA and HfrB strains and also the distance between these markers.
Introduction:
Genetic markers are the genes that are identifiable through
b.
To draw:
The order of the genes based on the given cotransduction data.
Introduction:
Cotransduction refers to the transfer of different bacterial genes together in one phage by transduction process.
c.
To determine:
The way an individual increases the chances of mapping gly gene relative to at least some of the other markers and the composition of the medium that would be used to map gly.
Introduction:
Tranductants refer to the cells resulting from the gene transfer. Transduction process is mainly mediated by bacteriophages.
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Genetics: From Genes To Genomes (6th International Edition)
- Conjugating E. coli was discovered to have an F factor aka Fertility factor. Explain what type of biomolecule F factor is and what features the F factor has which enhances its uptake and overall genetic recombination in E. coli. Also, in your answer detail the processes of F- to F+ conversion, F+ to HFR conversion and HFR to F' formation.arrow_forwardIf an E. coli auxotroph X could grow only on a medium containing leucine, and an auxotroph Y could grow only on a medium containing threonine. (i) Write the genotype of both E. coli strains. (ii) How would you test whether DNA from X could transform Y?arrow_forwardIn Escherichia coli, four different Hfr strains, derived from the same F* strains, were mated with F strains auxotrophic for a number of nutritional requirements (Arg Bio Cys Trp Gal His Lac Mal Xyl Leu Met). Matings were interrupted at various intervals and cells were plated on minimal medium supplemented with particular nutrients to test for gene transfer. The following results show the time of entry for each of the genes in four different Hfr strains. Table 1: Time-of-Entry Mapping Data* Hfr strains Genes lac" his" arg' bio 9. Cys gal" trp' 5 11.5 2.5 mal" xyl" leu met Hfr 1 Hfr 2 Hfr 3 Hfr 4 6.5 3.5 11 15 15 4 6. 17.5 5 14.5 3 20 * The numbers denote the number of minutes elapsed before a gene enters the F cells. Draw a circular map of E. coli chromosome.arrow_forward
- Time mapping is performed in a cross involving the genes his, leu, mal, and xyl. The recipient cells were auxotrophic for all four genes. After 25 minutes, mating was interrupted with the following results in recipient cells. Diagram the positions of these genes relative to the origin (O) of the F factor and to one another. (a) 90% were xyl+ (b) 80% were mal+ (c) 20% were his+ (d) none were leu+arrow_forwardIn a cotransformation experiment (see question 4 of More GeneticTIPS), DNA was isolated from a donor strain that was proA+ andstrC+ and sensitive to tetracycline. (The proA and strC genes conferthe ability to synthesize proline and confer streptomycin resistance,respectively.) A recipient strain is proA− and strC− and isresistant to tetracycline. After transformation, the bacteria werefirst streaked on a medium containing proline, streptomycin, andtetracycline. Colonies were then restreaked on a medium containingstreptomycin and tetracycline. (Note: Each type of medium hadcarbon and nitrogen sources for growth.) The following resultswere obtained:70 colonies grew on the medium containing proline, streptomycin,and tetracycline, but only 2 of these 70 colonies grew whenrestreaked on the medium containing streptomycin and tetracyclinebut lacking proline. If we assume the average size of the DNA fragments is 2 minutes,how far apart are these two genes?arrow_forwardFor each of the E. coli strains containing the lacoperon alleles listed, indicate whether the strain isinducible, constitutive, or unable to expressβ-galactosidase and permease.a. I+ o+ Z− Y+/ I+ ocZ+ Y+b. I+ o+ Z+ Y+/ I− ocZ+ Y−c. I+ o+ Z− Y+/ I− ocZ+ Y−d. I−P− o+ Z+ Y−/ I+ P+ ocZ− Y+e. Iso+ Z+ Y+/ I− o+ Z+ Y−arrow_forward
- In E. coli, four Hfr strains donate the following markers,shown in the order donated:Strain 1: M Z X W CStrain 2: L A N C WStrain 3: A L B R UStrain 4: Z M U R BAll these Hfr strains are derived from the same F+ strain.What is the order of these markers on the circularchromosome of the original F+?arrow_forwardYour TA gives you an Escherichia coli strain (AmreB) that carries a gene deletion in the mreB gene. As a result, the strain is not able to produce the MreB protein. Your task is to compare the morphology of actively growing AmreB cells to the parental wildtype strain (produces MreB). What difference in morphology will you likely observe? Would you expect the same in Staphylococcus aureus?arrow_forwardFour Hfr strains (A, B, C, D), were all derived from a F+ strain of E. coli. When each strain is used as a donor in an interrupted-conjugation experiment, the entry times of the first five markers are indicated in the table below (time of entry in minutes is indicated in brackets). Below is a partial map of the F+ strain (distances not proportional). Match the position of all genes with their correct position on this map. Strain A ala+ (5) ser+ (10) ade+ (15) his+ (25) val+ (30) 3 5 2 7 4 1 6 Strain B ade+ (12) his+ (22) val+ (27) pro+ (42) met+ (45) > sert ade+- 1. lac+ his+ ala+ met+ mal+ pro+ val+ Strain C pro+ (1) met+ (6) lac+ (11) mal+ (21) ala+ (31) 7. Strain D met+ (10) pro+ (15) val+ (30) his+ (35) ade+ (45) 2. 6. 3. 5. 4. 1. position #1 2. position #2 3. position #3 4. position #4 5. position #5 6. position #6 7. position #7arrow_forward
- Three different Hfr strains, derived from the same F* strains (H), were mated with F strains auxotrophic for a number of nutritional requirements (trp bio leu azi mal tyr gal ). Matings were interrupted at various intervals and cells were plated on minimal medium supplemented with particular nutrients to test for gene transfer. The following mapping data was obtained: Genes mal Hfr strains leu 25 trp 17 bio azi gal* tyr Hfr H Hfr 1 Hfr 2 Hfr 3 Hfr 4 4 10 10 18 3 9. 16 14 22 15 3 19 19 11 4 8 Draw a circular map of E. coli chromosome. Include distances (in minutes) between genes in your map and label arrowheads to show the origin of transfer for the five Hfr strains.arrow_forwardWhat is the simplest explanation for why patients have been identified with only one copy of the phosphofructokinase-1 gene (heterozygous), but no patients have been identified that lack both copies of the phosphofructokinase-1 gene (homozygous)? Patients lacking both copies of the phosphofructokinase-1 genes will be found once DNA sequencing technology can sequence whole genomes. Phosphofructokinase-1 is needed for nitrogen metabolism and there are no enzymes to replace this function, so cells die from ammonia toxicity. Phosphofructokinase-1 is a required enzyme for carbohydrate metabolism in all living cells, complete loss of this enzyme would be lethal. There are 6 phosphofructokinase-1 paralogous genes in humans and it is impossible to lack both type 1 copies when there are also types 4, 5, and 6.arrow_forwardA series of auxotrophic mutants were isolated in Neurospora. Examination of fungi containing these mutations revealed that they grew on minimal medium to which various compounds (A, B, C, D) were added; growth responses to each of the four compounds are presented in the following table. Give the order of compounds A, B, C, and D in a biochemical pathway. Outline a biochemical pathway that includes these four compounds and indicate which step in the pathway is affected by each of the mutations. Compound Mutation number А в С A D 134 276 987 773 772 146 333 123 + + I +arrow_forward
- Biology: The Dynamic Science (MindTap Course List)BiologyISBN:9781305389892Author:Peter J. Russell, Paul E. Hertz, Beverly McMillanPublisher:Cengage Learning